QCE Vault / Specialist Maths Vectors in two and three dimensions — Question 189 Original QCE Vault practice · 5 marks
Browse all questions Vectors revision Original practice exam Report an issue Q189 · Practice question Technology-free Complex familiar 5 marks
QUESTION 189 (5 marks) A triangular panel has vertices A = ( 1 , 0 , 0 ) A=(1,0,0) A = ( 1 , 0 , 0 ) , B = ( 0 , 2 , 0 ) B=(0,2,0) B = ( 0 , 2 , 0 ) and C = ( 0 , 0 , 3 ) C=(0,0,3) C = ( 0 , 0 , 3 ) . a) Determine a Cartesian equation of its plane.
[3 marks] b) Determine the exact area of the panel.
[2 marks] WORKED SOLUTION
Practice marking scheme 5 marks ANSWER (a) 6 x + 3 y + 2 z = 6 6x+3y+2z=6 6 x + 3 y + 2 z = 6 . (b) 7 / 2 7/2 7/2 square units. Worked solution
(a) The edge vectors are ( − 1 , 2 , 0 ) (-1,2,0) ( − 1 , 2 , 0 ) and ( − 1 , 0 , 3 ) (-1,0,3) ( − 1 , 0 , 3 ) . Their cross product is ( 6 , 3 , 2 ) (6,3,2) ( 6 , 3 , 2 ) , a normal. Through A the equation is 6 ( x − 1 ) + 3 y + 2 z = 0 6(x-1)+3y+2z=0 6 ( x − 1 ) + 3 y + 2 z = 0 . All three vertices satisfy it. (b) The triangle area is half the cross-product magnitude: 1 2 36 + 9 + 4 = 7 / 2 \frac12\sqrt{36+9+4}=7/2 2 1 36 + 9 + 4 = 7/2 . The full magnitude would be the parallelogram area. Equivalent mathematically justified methods accepted.
Exact answers unless a decimal accuracy is specified.
Part a: Form two independent edges.
[1 mark] Part a: Compute their cross product.
[1 mark] Part a: Use a vertex to obtain the plane.
[1 mark] Part b: Use half the cross-product magnitude.
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